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Can anyone show the way to get BC

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Manager
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Can anyone show the way to get BC [#permalink] New post 18 Aug 2003, 11:57
Can anyone show the way to get BC?
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answer [#permalink] New post 18 Aug 2003, 13:01
sure man.

First, let's call the point at the right angle "D", so BD + DC = BC

Well, since angle B is 60, then angle A is 30, and the side across from the 30 degree angle in a 30-60-90 triange is always half of the hypoteneuse. So that makes BD = .5

Now, the longer leg in a 30-60-90 triangle is sqrt(3) times the shorter leg, so AD is .5 * sqrt(3). Because triangle ADC is a 45-45-90 triangle, then AD = DC = .5 * sqrt(3). so BC is .5 + .5 * sqrt(3).

I think column B is bigger.
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 [#permalink] New post 30 Aug 2003, 23:50
Does anyone has an alternative way to derive the answer in attached file?

By the way, is it confirmed that the side across from the 30 degree angle in a 30-60-90 triange is always half of the hypoteneuse?
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 [#permalink] New post 31 Aug 2003, 19:06
I did it in the same way as mciatto. (30-60-90 and 45-45-90 triangles)

However unless i'm missing something wouldn't this make Column A bigger?
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 [#permalink] New post 01 Sep 2003, 05:38
I did not know 30-60-90 rule exists.

If it is true, the answer is correct. Try again, or you can post your workings here.
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 [#permalink] New post 01 Sep 2003, 10:31
Yeah your right, i made a stupid mistake.
  [#permalink] 01 Sep 2003, 10:31
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